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This is a follow-up post to a previous question of mine:

Check whether numpy array row is smaller than the next

Suppose i have the following numpy array:

a=np.reshape(np.array([[79,np.nan,87,77,92,133,99,121,103,118,126, 
133,131,67]]),(7,2))

In [1]: a
Out[1]: 
array([[  79.,   nan],
       [  87.,   77.],
       [  92.,  133.],
       [  99.,  121.],
       [ 103.,  118.],
       [ 126.,  133.],
       [ 131.,  67.]])

I would like to create a new column or array which will be a True/False indicator testing the following proposition:

a[-1, 0] < a[1:, 0] and a[-1, 1] > a[1:, 1]

The result that i expect is the following:

False (because the first value of column 1 is nan)
False
True
True
False
True
False

I have tried different variations of the solutions described in my previous post, but so far i have been unsuccessful.

EDIT:

The idea is to test whether 87<92 and at the same time 77>133 which is False. Then 92<99 and 133>121 which is True etc.

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  • Your result will be a vector 6 times False because 131 (a[-1,0]) is > every other value in that column. Same for col 2. Can you please explain your proposition more exactly? Commented Feb 27, 2018 at 22:43
  • I just did, thank you. Hopefully this is more clear now. Commented Feb 27, 2018 at 22:49

1 Answer 1

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You will use exactly the same strategy than Tai's answer in your previous post:

b = np.diff(a, axis=0)

[[   8.   nan]
 [   5.   56.]
 [   7.  -12.]
 [   4.   -3.]
 [  23.   15.]
 [   5.  -66.]]

And now you use a the logic function np.logical_and on array b. Since you always want a trailing False, we can just append it:

result = np.logical_and(b[:,0] > 0, b[:,1] < 0)
result = np.append(result, np.array([False]))

[False False  True  True False True False]

(edited my post according to your comments)

Note: Comparing values with nan will always return False. So if nan appears in some row in the middle, it will always yield two rows that evaluate to False.

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7 Comments

Just to break it down: The first False = 79 < 87 and nan > 76, second False = 87 < 92 and 77 > 133, third True = 92 < 99 and 133 > 121, fourth True = 99 < 103 and 121 > 118, fifth False = 103 < 126 and 118 > 133, sixth False = 126 < 131 and 133 > 672. That's what my code does and what your description suggests. If that's wrong, please tell for every row how the values should be calculated. Especially since you build differences from a 7-row-array between neighboring rows and still have 7 results instead of 6.
You are in fact correct. My apologies, i made a mistake in the last row. The value should be 67 not 672. I corrected that now. And the last (7nth row) is always false because there is not another row to compare it with.
One more thing. Can i test the same proposition using assert? It appears that with assert in Numpy you can only test certain things.
The assert functions in numpy.testing are intended for testing. So in a test suite, you could say numpy.testing.assert_equal(result, np.array([False,False, True, True, False, True, False])) which would raise an error if not all elements are equal.
Thank you, that's really helpful. I would like to have an if statement to take action is assert fails. Should i do this using a combination of try and exception? Or is there a better way?
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